If barometric pressure on a mountain top is 735 mm of mercury, then the pressure (in N/m2) will be:
98060
Barometric Pressure Calculation
Barometric pressure refers to the pressure exerted by the atmosphere at a given point. It is often measured in millimeters of mercury (mm Hg) or inches of mercury (in Hg). To perform calculations in the International System of Units (SI), we need to convert this pressure into Newtons per square meter ($\text{N/m}^2$), which is also known as Pascals (Pa). This conversion is fundamental in physics and engineering problems related to fluids.
Pressure Conversion Fundamentals
The pressure exerted by a column of fluid can be calculated using the formula:
$\text{P} = \rho \text{gh}$
Where:
- $\text{P}$ is the pressure in Pascals ($\text{N/m}^2$).
- $\rho$ (rho) is the density of the fluid (in this case, mercury) in kilograms per cubic meter ($\text{kg/m}^3$).
- $\text{g}$ is the acceleration due to gravity in meters per second squared ($\text{m/s}^2$).
- $\text{h}$ is the height of the fluid column in meters (m).
Given Values and Unit Preparation
Let's identify the values provided in the question and the standard physical constants required for the calculation. We must ensure all values are in consistent SI units before proceeding.
| Parameter | Value (Given/Standard) | Unit Conversion (if needed) |
|---|---|---|
| Barometric Pressure Height ($\text{h}$) | $735 \text{ mm}$ of mercury | $735 \text{ mm} = 0.735 \text{ m}$ (since $1 \text{ m} = 1000 \text{ mm}$) |
| Density of Mercury ($\rho$) | $13600 \text{ kg/m}^3$ | No conversion needed (already in SI units) |
| Acceleration due to Gravity ($\text{g}$) | $9.81 \text{ m/s}^2$ | No conversion needed (already in SI units) |
Note: While $9.8 \text{ m/s}^2$ is a common approximation for $\text{g}$, using $9.81 \text{ m/s}^2$ often yields results closer to options in competitive exams. We will use $9.81 \text{ m/s}^2$ for a more precise calculation.
Pressure Calculation Steps
Now, we will substitute these values into the pressure formula $\text{P} = \rho \text{gh}$ to find the pressure in $\text{N/m}^2$.
-
Substitute the values:
$\text{P} = (13600 \text{ kg/m}^3) \times (9.81 \text{ m/s}^2) \times (0.735 \text{ m})$ -
Perform the multiplication:
$\text{P} = 133416 \text{ kg/(m} \cdot \text{s}^2) \times 0.735 \text{ m}$
$\text{P} = 98010.76 \text{ kg/(s}^2 \cdot \text{m})$ -
Express in correct units:
Since $1 \text{ kg/(s}^2 \cdot \text{m}) = 1 \text{ N/m}^2$ (or $1 \text{ Pa}$), the pressure is:
$\text{P} = 98010.76 \text{ N/m}^2$
Comparing Results with Options
Our calculated pressure is approximately $98010.76 \text{ N/m}^2$. Let's compare this with the given options:
- Option 1: $980.60 \text{ N/m}^2$
- Option 2: $98060 \text{ N/m}^2$
- Option 3: $9806.0 \text{ N/m}^2$
- Option 4: $98.060 \text{ N/m}^2$
The calculated value $98010.76 \text{ N/m}^2$ is extremely close to option 2, which is $98060 \text{ N/m}^2$. The slight difference is likely due to rounding of constants (like $\text{g}$ or $\rho$) used in the problem's original calculation or rounding of the final answer. In multiple-choice questions, we select the closest available option.
Therefore, the pressure will be approximately $98060 \text{ N/m}^2$.
Important Questions from Hydrostatic Law
What is the water pressure on the sea bottom at a depth of 6500 m? The specific gravity of sea water is assumed to be 1.03.
Which law states that the rate of increase of pressure in a vertical direction is equal to weight density of a fluid at a point?
A beaker contains water up to a height of h/3. Find the location of centre of pressure.
The specific weight W of ocean water changes according to the equation W = Wo + C√h. W0 = Specific weight of ocean water at free surface and C is a constant. The pressure as a function of depth h is given by
A rectangular plane surface that lies in a vertical plane in water is 2 m wide and 3 m deep. Find the Total pressure if the upper edge is horizontal and 2 m below the free water surface.
